Let $I(x) = \frac{1}{2\pi} \int_{2}^2 \sqrt{4y^2}\lnxydy$. Then $I(x)$ is a concave function and
\begin{equation}
I(x)=
\begin{cases}
\frac{1}{4}x^2\frac{1}{2}, &\text{if } x\leq2 \\
\frac{1}{4}x^2\frac{1}{2}\frac{1}{4}x\sqrt{x^24}+\ln \frac{x+\sqrt{x^24}}{2} &\text{if } x>2
\end{cases}
\end{equation}
Let $J(x)$ be the Legendre transform of $\frac{1}{2}x^2  I(x)$, i.e. $J(x): = \sup_{y\in \mathbb{R}}\{xy  \frac{1}{2}y^2 + I(y)\}$, then $J(x)$ has a quite simple form
\begin{equation}
J(x) =
\begin{cases}
x^2\frac{1}{2} &\text{if } x\leq1 \\
\frac{1}{2}x^2 + \ln x &\text{if } x>1
\end{cases}
\end{equation}
The simple form of $J(x)$ suggests some hidden deeper truth. My question is, can we find $J(x)$ directly without calculating $I(x)$ explicitly?

10$\begingroup$ for $y>2$, the function $y^2/2I(y)$ is precisely the rate function for the top eigenvalue of GOE/GUE, up to a constant, see section 6 of link.springer.com/content/pdf/10.1007/PL00008774.pdf. Thus you are asking about the logmgf of the top eigenvalue, which in turns can be computed using the Selberg integral. I am however not sure this qualifies as a simpler, or deeper, truth.. $\endgroup$– ofer zeitouniCommented Jun 13, 2020 at 16:24
1 Answer
Claim. $J(x)=\tfrac12x^2+\lnx+c$ for $x>1$ with $c$ constant.
Proof: Here, an explicit form for $I(x)$ is not needed but I can't use it to prove $c=0$.
For $x>1$, the substitution $y=2\cos t$ followed by $(x\cos t)(x\cos s)=x^21$ yields $$I'(2x)=\frac1{2\pi}\int_{2}^2\frac{\sqrt{4y^2}}{2xy}\,dy=x\frac1\pi\int_0^\pi\frac{x^21}{x\cos t}dt=x\sqrt{x^21}$$ where $'=d/dx$. Thus $I'(x)+1/I'(x)=x$ for $x>2$.
We have $J(x)=xy_*\tfrac12y_*^2+I(y_*)$ where $xy_*+I'(y_*)=0$. This is equivalent to $y_*x+\frac1{y_*x}=y_*$ using the above identity, which implies $y_*=x+\frac1x$. Thus $J(x)=\frac12x^2\frac1{2x^2}+I(x+\frac1x)$ so we now evaluate $I(x+\frac1x)$.
Let $K(x)=xI^*(x+\frac1x)$ where $^*=d/d(x+\frac1x)$ so the above identity simplifies to $K(x)^2(x^2+1)K(x)+x^2=0$. By inspection, we have $K(x)=1$ or $x^2$; however, if $K(x)=x^2$ then $I'$ behaves like $x$ asymptotically, which is a contradiction. Thus $K(x)=1$, so $I^*(x+\frac1x)=\frac1x$ implies $I'(x+\frac1x)=\frac1x(1\frac1{x^2})$ or that $I(x+\frac1x)=\frac1{2x^2}+\lnx+c$. This holds for $x>1$ so $J(x)=\tfrac12x^2+\lnx+c$ for $x>1$.

$\begingroup$ $I(2)=1/2$ proves $c=0$ but that requires direct evaluation. $\endgroup$ Commented Mar 3 at 13:18